2022年安徽省淮南市高考数学一模试卷（理科）（附答案详解）

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2022
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14•••AC2=AB2+BC2-2AB-BC-cos÷4BC=3,.•MC=V3..-.BC2+AC2=AB2,-.^ACB=^,^ACD=^,2.61•sinZ.ACD=2}øù4BCDḄ☢úS=SUBC+S“cD=-BC+^AC-CD-sin÷4CD=û+f=3V34(2)ᙠUCOAD2=AC2+DC2-2AC-DC-cos^ACD=1,AD=1,sinD=—•smZ-ACD=—.AD2ூ(᪆௃(1)üý᳛ḄᦪúijN4BC=ᑭᵨÿ
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ᳮ
ᳮḄᵨ!Ạ⚪.18.ூ$ᫀ௃'(1)᪷*⚪+,“ᵬ/010ᐗஹ1ᐗஹ2ᐗஹ”156A,829:;<=>?p(4)=0.5,p(4)=0.2,p(4)=0.3,3@ᳮ,“A/010ᐗஹ1ᐗஹ2ᐗ”156BB,29:;<=>?p(B])=0.4,p(B)=0.4,p(fi)=0.2,23ᵫ⚪D56A,&F56BB,GH=IJ,ᵬKA/0L156M,22Mljᨵ'M=+AB>32O'P(M)=P(712)P(BI)+PQ43)P(BI)+PQ43)P(B2)=0.2X0.4+0.3x0.4+0.3X0.4=0.32,ᦑᵬKA/0LḄᭆ᳛1'0.32.(2)ᵫ⚪D6ḄTUV1'0,1,2,3,4,ᑣᨵ'p(X=0)=0.5x0.4=0.2,p(f=1)=0.5X0.4+0.2x0.4=0.28,=2)=0.5x0.2+0.3x0.4+0.2x0.4=0.3,=3)=0.2X0.2+0.3x0.4=0.16,P(f=4)=0.3x0.2=0.06,ᡠfḄᑖ\ᑡ1'

1501234P0.20.280.30.160.06ᦑXḄᦪ^_`'F(0=0X0.2+1X0.28+2X0.3+3x0.16+4X0.06=1.6.ூ᪆௃(1)᪷*mDᩩ6oᔠH=IJ56Ḅᭆ᳛qr8sᓽ.(2)ᵫ⚪DfḄTUV1'0,1,2,3,4,ᑖvwḄᭆ᳛xoᔠ_`8s,ᓽ.⚪y⌕ὃ{ᦣ}~ᑖ\ᑡḄὃ_`8sḄᵨ!Ạ⚪.19.ூ$ᫀ௃'(1)Bn=l,%=2—2%,ᑠ=|,Bn>2Qig…%i=2-2a?aa=2-2aH◀Oᓽ=(2);nr2᦮ᳮ1'-----=1ᜐ"-=---1(712),•.aTl-al-aஹ¢71nni1ᓽ£¤¥=1522),§ᓝ©31✌⚗8¬11Ḅ¬ᦪᑡ®1-anᡠᓝ=3+(-1)=71+2,1¯ᦑan=^S€N*).'(2)ᵫ(1)Oµ®="+2,᦮ᳮ1'an=^(n€N*)®ᵫᩩ6O"=Ḅ஺2…ᓽ=·...2=_j_>_i_=}2_____nTn(n+2)2(n+2)(n+3)kn+2n+3'•-=Tf+r/...2>(1-1..._1___1_)i__1__|),+r4++=4(1cஹn+222:-5>--------=---4nn+33n+13fᡠBneN*is>a-1.nn+1ூ᪆௃(1)ÁÂᑭᵨᦪᑡḄ⌴ÄᐵÆᦪᑡ1¬ᦪᑡÇÈᦪᑡḄÉ⚗8s®(2)ᑭᵨÊËrÌÍ⚗HÎrḄᵨᦪᑡḄÌÇÈoÏ.Ð16⚓ᐳ19⚓

16⚪ὃḄDÓ⌕Ô'ᦪᑡḄᐵÆsḄᣚ¬ᦪᑡḄ
ÖÊËrÌÍ⚗HÎrḄÌy⌕ὃ^×ḄØÙTÚÌᦪ^ÛÜTÚÝ᫏⚪.20.ூ$ᫀ௃'(1)Ὂ.à=\2,ᵯ.(â—ᵯ)=0,ã•ä=0,âj,åæÔP(2,&)ᙠèᙊCê•ës-2,0),ë2(2,0),?ᵫèᙊ
ÖO2a=IP&I+IPF2I=J(-2-2)2+(0-V2)2+J(2-2)2+(0-V2)2=4V2a=2V2,b2=a2-4=4,ᑣèᙊCḄ᪗íîï1ð+=1?84(2)Ꮇòóᙠ
ÔMôõ⌕ö1Á÷1ø᳛¤1ùᡠòÁ÷¢%=my4-2,M(x0),i4(x,y),B(x,y)0,1122úÁ÷lḄîïû=ü/+206/?)ýᐭèᙊîïs+^=1,᦮ᳮO(ÿ+2)y2+844my—4=0,4m4ᨴ+2=%2=•஺ᑖ~18,../£"4+/8=0,ᓽ+᧡=°..._yi_+__o=myi+2-Xomy+2-x20᦮ᳮ2myy+(2-3+%)=஺12•**2m(----^ߟ)+(2—%o)(—)=0,'7712+27'U?712+27.%m(4-%)=0,ᵫAx=4,00ᡠ!ᙠM(4,0)#$⌕&.ூ)᪆௃(1)-./•.0=1(230j_45ᔠ7p(2,8)ᙠ9ᙊC;&)a,b,ᑮ9ᙊCḄ᪗?@A.(2)BCD%=my4-2,M(xiO),4(%iyi),8(E/2)FCD1Ḅ@AG=my4-2(m6OR)Hᐭ9ᙊ@A1+4=1ᑭᵨLMNᳮ5ᔠkMA+kM8=0,Oᓄ&)ᓽ1.84Q⚪ὃTCDU9ᙊḄVWᐵYḄZᔠ[ᵨὃTOᓄ\]^_`abcd᫏⚪.

1721.ூfᫀ௃):(l)y=஻x)ḄNjkl(0,l)U(l,+8)ml/஺)=nᡠ♦஺)=o;q9(%)=1-Inx,ᑣg'(%)=sᡠtᦪy=g(x)ᙠvw(0,1)ᓫyᙠvw(1,+8)ᓫz.{mlg(l)=Oᡠ|}£(0,1)11(1,+8)0஺)<0,/(%)<0ᡠtᦪy=/(%)ᙠvw(0,1),(1,+8);ᙳᓫ⌴z(2),,Ax2—Ax>(eAx—l)lnx(x—>(e^—l)lnx|ᐭ>0,%>1G-1>0,ᡠ&1ᓄl;ᜐeAX-lx-1B|J/(eAx)>/(%),vA>0᧕6(l,+oo),ᵫ(1),y=/(%)ᙠ(1,+8)ᓫ⌴z,ᦑ◤e&W%ᙠ(1,+oo);aᡂ.ᯠᦪீᓽ4S?ᙠ(1,+8);aᡂ.1-lnxq8(x)>1),|J"(X)=x2|€(l.e)(pf(x)>0,tᦪy=?(x)ᓫ⌴y|G6(e,+8)(p0,ᦑ4Ḅ¡c(0,3.ூ)᪆௃(1)ᵫ((¢=£ἠ.qg(x)=l-:)%,ᑭᵨg(x)Ḅ¥ᦪ1/(x)Ḅᓫvw(2)1ᓄlᗞ>¨,/(e^)>f(x),©ª4<«ᙠ(1,+8);aᡂ.q(p(x)=e*x—1xᑭᵨ¥ᦪ1&;IḄ¡.Q⚪ὃTtᦪḄᓫឋᨬ°⚪ὃT¥ᦪḄ[ᵨ±²⚪.(X=t222.ூfᫀ௃)(1)ᵫ(tl³ᦪ)(tl³ᦪ),´otµDCḄ¶-@Aly2=4%.4(2)ᵫ2cosJ-sin஺=72x-y=4,y2=4%ὶ2x—y=4b(1,2),8(4,4),¸18⚓ᐳ19⚓

18ᡠ4Bd7ᙶ᪗l\AB\=V45,ᦑABlCÁḄᙊḄCÂᙶ᪗@Al(x-Ä+(y-I)2=Èᓽx2+y2-5x-2y-4=0,Wx=pcosd,y=psind,Hᐭpz-5pcosx—2psin6-4=0.ூ)᪆௃(1)CÎᑭᵨOᣚᐵYᙠ³ᦪ@A᩽ᙶ᪗@AÑCÂᙶ᪗@AwÒÓOᣚ(2)ᑭᵨ@AÔḄ)Õ&35Ö.Q⚪ὃTḄ×⌕7³ᦪ@A᩽ᙶ᪗@AÑCÂᙶ᪗@AwḄOᣚ@AÔḄ)ÕØ⌕ὃTÙḄÚ`abÑᦪ\Ûab±ÜẠ⚪.23.ூfᫀ௃)(1)||2x+>Jm\—\2x—1||<\2x+y/m—(2x—1)|=Vm+1,ᦑi—1S|2x+ᑗᔣ—|2x—1|W+1,A—y/m—1=-2,m=1.ফᵫপ1è+&=1,ᡠa2+b2=(a2+2)+(b2+i)—3=[(a2+2)+(/+i)].(&+e)ï3=(2+ð+▢)-324-3=1,'b2+la2+2y|òó|*=3ᓽa=0,b=±löᡂᦑ஺2+᮱Ḅᨬøl1.a<+22ூ)᪆௃(1)ᑭᵨùḄúûüj&)tᦪḄkᑭᵨᨬø&)mᓽ1.(2)ᑭᵨ(1)Ḅ5Ö-.0ýÕ5ᔠÜQ&)⊤MḄᨬøᓽ1.Q⚪ὃTtᦪḄᨬḄ&ÕÜQḄ[ᵨὃTOᓄ\]^_`abcd᫏⚪.